Question

The variance of first n odd natural numbers is $\frac{n^2 - 1}{3}$. The sum of first n odd natural number is $n^2$ and the sum of square of first n odd natural numbers is $\frac{ n (4n^2 + 1)}{3}$ .

• A. Statement 1 is true, Statement 2 is false
• B. Statement 1 is true, Statement 2 is true; Statement 2 is not a correct explanation for Statement 1
• C. Statement 1 is false, Statement 2 is true
• D. Statement 1 is true, Statement 2 is true, Statement 2 is a correct explanation for Statement 1

Answer 1

The Correct Option is A

The correct option is A, Statement 2: Sum of first n odd natural numbers is not equal to \(n^2\). So, statement - 2 is false.

Answer 2

1. Sum of the first \( n \) odd natural numbers:
The \( n \)th odd number is \( 2n-1 \). The sum of the first \( n \) odd numbers is:
\[ 1 + 3 + 5 + \cdots + (2n-1) \]
This sum is known to be:
\[ S_n = n^2 \]
So, the statement "Sum of the first \( n \) odd natural numbers is equal to \( n^2 \)" is true.
2. Sum of the squares of the first \( n \) odd natural numbers:
Given that:
\[ \text{Sum of squares of the first } n \text{ odd natural numbers} = \frac{n(4n^2 - 1)}{3} \]
3. Variance of the first \( n \) odd natural numbers:
Variance \( \sigma^2 \) is calculated using the formula:
\[ \sigma^2 = \frac{1}{n} \left( \sum_{i=1}^{n} x_i^2 \right) - \left( \frac{1}{n} \sum_{i=1}^{n} x_i \right)^2 \]
Here, \( x_i \) represents the \( i \)th odd number.
Given:
\[ \sum_{i=1}^{n} x_i = n^2 \]
\[ \sum_{i=1}^{n} x_i^2 = \frac{n(4n^2 - 1)}{3} \]
Now, let's compute the variance:
First, find the mean (\( \mu \)):
\[ \mu = \frac{1}{n} \sum_{i=1}^{n} x_i = \frac{n^2}{n} = n \]
Next, compute the variance:
\[ \sigma^2 = \frac{1}{n} \left( \sum_{i=1}^{n} x_i^2 \right) - \mu^2 \]
\[ \sigma^2 = \frac{1}{n} \left( \frac{n(4n^2 - 1)}{3} \right) - n^2 \]
\[ \sigma^2 = \frac{4n^2 - 1}{3} - n^2 \]
\[ \sigma^2 = \frac{4n^2 - 1}{3} - \frac{3n^2}{3} \]
\[ \sigma^2 = \frac{4n^2 - 1 - 3n^2}{3} \]
\[ \sigma^2 = \frac{n^2 - 1}{3} \]
This matches the given variance formula \( \frac{n^2 - 1}{3} \).
4. Verification of the statements:
- Statement 1: The sum of the first \( n \) odd natural numbers is \( n^2 \). (This is true)
- Statement 2: The sum of the first \( n \) odd natural numbers is not equal to \( n^2 \). (This is false)
Therefore, the correct option is that Statement 2 is false, and we have verified the variance formula as well.